\(\int \frac {(d+e x)^2}{a^2+2 a b x+b^2 x^2} \, dx\) [1508]

Optimal result

Integrand size = 26, antiderivative size = 51 \[ \int \frac {(d+e x)^2}{a^2+2 a b x+b^2 x^2} \, dx=\frac {e^2 x}{b^2}-\frac {(b d-a e)^2}{b^3 (a+b x)}+\frac {2 e (b d-a e) \log (a+b x)}{b^3} \]

[Out]

e^2*x/b^2-(-a*e+b*d)^2/b^3/(b*x+a)+2*e*(-a*e+b*d)*ln(b*x+a)/b^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {27, 45} \[ \int \frac {(d+e x)^2}{a^2+2 a b x+b^2 x^2} \, dx=-\frac {(b d-a e)^2}{b^3 (a+b x)}+\frac {2 e (b d-a e) \log (a+b x)}{b^3}+\frac {e^2 x}{b^2} \]

[In]

Int[(d + e*x)^2/(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

(e^2*x)/b^2 - (b*d - a*e)^2/(b^3*(a + b*x)) + (2*e*(b*d - a*e)*Log[a + b*x])/b^3

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \frac {(d+e x)^2}{(a+b x)^2} \, dx \\ & = \int \left (\frac {e^2}{b^2}+\frac {(b d-a e)^2}{b^2 (a+b x)^2}+\frac {2 e (b d-a e)}{b^2 (a+b x)}\right ) \, dx \\ & = \frac {e^2 x}{b^2}-\frac {(b d-a e)^2}{b^3 (a+b x)}+\frac {2 e (b d-a e) \log (a+b x)}{b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.92 \[ \int \frac {(d+e x)^2}{a^2+2 a b x+b^2 x^2} \, dx=\frac {b e^2 x-\frac {(b d-a e)^2}{a+b x}+2 e (b d-a e) \log (a+b x)}{b^3} \]

[In]

Integrate[(d + e*x)^2/(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

(b*e^2*x - (b*d - a*e)^2/(a + b*x) + 2*e*(b*d - a*e)*Log[a + b*x])/b^3

Maple [A] (verified)

Time = 2.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.24

method	result	size
default	\(\frac {e^{2} x}{b^{2}}-\frac {2 e \left (a e -b d \right ) \ln \left (b x +a \right )}{b^{3}}-\frac {a^{2} e^{2}-2 a b d e +b^{2} d^{2}}{b^{3} \left (b x +a \right )}\)	\(63\)
norman	\(\frac {\frac {e^{2} x^{2}}{b}-\frac {2 a^{2} e^{2}-2 a b d e +b^{2} d^{2}}{b^{3}}}{b x +a}-\frac {2 e \left (a e -b d \right ) \ln \left (b x +a \right )}{b^{3}}\)	\(68\)
risch	\(\frac {e^{2} x}{b^{2}}-\frac {2 e^{2} \ln \left (b x +a \right ) a}{b^{3}}+\frac {2 e \ln \left (b x +a \right ) d}{b^{2}}-\frac {a^{2} e^{2}}{b^{3} \left (b x +a \right )}+\frac {2 a d e}{b^{2} \left (b x +a \right )}-\frac {d^{2}}{b \left (b x +a \right )}\)	\(86\)
parallelrisch	\(-\frac {2 \ln \left (b x +a \right ) x a b \,e^{2}-2 \ln \left (b x +a \right ) x \,b^{2} d e -x^{2} b^{2} e^{2}+2 \ln \left (b x +a \right ) a^{2} e^{2}-2 \ln \left (b x +a \right ) a b d e +2 a^{2} e^{2}-2 a b d e +b^{2} d^{2}}{b^{3} \left (b x +a \right )}\)	\(100\)

[In]

int((e*x+d)^2/(b^2*x^2+2*a*b*x+a^2),x,method=_RETURNVERBOSE)

[Out]

e^2*x/b^2-2/b^3*e*(a*e-b*d)*ln(b*x+a)-1/b^3*(a^2*e^2-2*a*b*d*e+b^2*d^2)/(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.80 \[ \int \frac {(d+e x)^2}{a^2+2 a b x+b^2 x^2} \, dx=\frac {b^{2} e^{2} x^{2} + a b e^{2} x - b^{2} d^{2} + 2 \, a b d e - a^{2} e^{2} + 2 \, {\left (a b d e - a^{2} e^{2} + {\left (b^{2} d e - a b e^{2}\right )} x\right )} \log \left (b x + a\right )}{b^{4} x + a b^{3}} \]

[In]

integrate((e*x+d)^2/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")

[Out]

(b^2*e^2*x^2 + a*b*e^2*x - b^2*d^2 + 2*a*b*d*e - a^2*e^2 + 2*(a*b*d*e - a^2*e^2 + (b^2*d*e - a*b*e^2)*x)*log(b
*x + a))/(b^4*x + a*b^3)

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.18 \[ \int \frac {(d+e x)^2}{a^2+2 a b x+b^2 x^2} \, dx=\frac {- a^{2} e^{2} + 2 a b d e - b^{2} d^{2}}{a b^{3} + b^{4} x} + \frac {e^{2} x}{b^{2}} - \frac {2 e \left (a e - b d\right ) \log {\left (a + b x \right )}}{b^{3}} \]

[In]

integrate((e*x+d)**2/(b**2*x**2+2*a*b*x+a**2),x)

[Out]

(-a**2*e**2 + 2*a*b*d*e - b**2*d**2)/(a*b**3 + b**4*x) + e**2*x/b**2 - 2*e*(a*e - b*d)*log(a + b*x)/b**3

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.31 \[ \int \frac {(d+e x)^2}{a^2+2 a b x+b^2 x^2} \, dx=\frac {e^{2} x}{b^{2}} - \frac {b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}}{b^{4} x + a b^{3}} + \frac {2 \, {\left (b d e - a e^{2}\right )} \log \left (b x + a\right )}{b^{3}} \]

[In]

integrate((e*x+d)^2/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")

[Out]

e^2*x/b^2 - (b^2*d^2 - 2*a*b*d*e + a^2*e^2)/(b^4*x + a*b^3) + 2*(b*d*e - a*e^2)*log(b*x + a)/b^3

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.27 \[ \int \frac {(d+e x)^2}{a^2+2 a b x+b^2 x^2} \, dx=\frac {e^{2} x}{b^{2}} + \frac {2 \, {\left (b d e - a e^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} - \frac {b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}}{{\left (b x + a\right )} b^{3}} \]

[In]

integrate((e*x+d)^2/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")

[Out]

e^2*x/b^2 + 2*(b*d*e - a*e^2)*log(abs(b*x + a))/b^3 - (b^2*d^2 - 2*a*b*d*e + a^2*e^2)/((b*x + a)*b^3)

Mupad [B] (verification not implemented)

Time = 9.77 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.39 \[ \int \frac {(d+e x)^2}{a^2+2 a b x+b^2 x^2} \, dx=\frac {e^2\,x}{b^2}-\frac {a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2}{b\,\left (x\,b^3+a\,b^2\right )}-\frac {\ln \left (a+b\,x\right )\,\left (2\,a\,e^2-2\,b\,d\,e\right )}{b^3} \]

[In]

int((d + e*x)^2/(a^2 + b^2*x^2 + 2*a*b*x),x)

[Out]

(e^2*x)/b^2 - (a^2*e^2 + b^2*d^2 - 2*a*b*d*e)/(b*(a*b^2 + b^3*x)) - (log(a + b*x)*(2*a*e^2 - 2*b*d*e))/b^3